STM32F411RE I2C Write/Read an External EEPROM Memory 24LC1025

[rodrigoruiz1230]

Hello community :)

I need help using the EEPROM Memory 24LC1025.

I've this address for the 24LC1025 EEPROM ( A0 = 0, A1 = 0 and A2 = 1). And I'm trying to write/read on the data address range 10000h-1FFFFh but I'm not able to, when I try to do anything in this sector it rollovers to the data address range 0000h-FFFFh and I haven't figured out why.

Here's part of my code:

I2C_HandleTypeDef hi2c1;

uint8_t dataread[127];

int main(void)
{
  HAL_Init();
  SystemClock_Config();
  MX_GPIO_Init();
  MX_DMA_Init();
  MX_I2C1_Init();
  MX_USART6_UART_Init();

  HAL_I2C_Mem_Read(&hi2c1, 0x54<<1, 0x0000, I2C_MEMADD_SIZE_16BIT, dataread, sizeof(dataread), 1000);
  HAL_UART_Transmit(&huart6, dataread, sizeof(dataread), HAL_MAX_DELAY);

  while (1)
  {
  }
}

I've modified the Mem_Read function incresing MemAddress from uint16_t to uint32_t because I wasn't able to reach one of the higher addresses.

In the datasheet the method of accesing different blocks of memory or different sectors is by modifying (setting or clearing) a "bit block", is there a way of doing this using the HAL? Where should I look to modify the content of the bytes sent to the I2C device to set or clear the bit block?

[TDK · Accepted solution]

The 16th bit of the address is B0 which is part of the slave address.

To write to 0x10000 and above, use the slave address 0b1010100 << 1.

To write below that, use the slave address 0b1010000 << 1.

[Karl Yamashita]

You didn't need to modify the HAL_I2C_Mem_Read function as 16bit address works.

You need to read a little bit more about the Block bit B0. It selects the 2 blocks that are available on that EEPROM. 

If B0 = 0, then you are accessing address 0-0xFFFF. 
If B0 = 1, then you are accessing address 10000-1FFFF. So if you pass 0x800 for the memory address, you're actually reading address 0x10800.

If B0 = 0, then control byte (slave address) is 0x50.

If B0 = 1, then control byte (slave address) is 0x54

[rodrigoruiz1230]

Thank you so much for your help. :) @TDK 

[rodrigoruiz1230]

Thank you very much for your help @Karl Yamashita. Also your solution is very accurate and very informative. :)